各类平均与数列极限

已知数列$\{a_n\},\{b_n\}$满足$a_0=a,b_0=b(a>0,b>0,a\neq b)$,且
\[\begin{cases}
a_{n+1}=\frac{1}{2}(a_n+b_n)\\
b_{n+1}=\sqrt{a_{n+1}b_n}.
\end{cases}(n=0,1,2,\ldots)\]
试求$\{a_n\},\{b_n\}$的通项公式以及它们的极限.

由题设条件有
\[\left( \frac{a_{n+1}}{b_{n+1}} \right) ^2=\frac{a_{n+1}^{2}}{a_{n+1}b_n}=\frac{a_{n+1}}{b_n}=\frac{1}{2}\left( \frac{a_n}{b_n}+1 \right).\]
令$t_n=\frac{a_n}{b_n} (n=0,1,2,\ldots)$,则
\[t_{n+1}^2=\frac{1}{2}(t_n+1),\quad t_0=\frac{a}{b}.\]
再令$t_n=\frac{1}{2}x_n$,则
\[x_{n+1}^2=x_n+2,\quad x_0=2\frac{a}{b}.\]
由方程$x_0=t+\frac{1}{t}=2\frac{a}{b}$,解得$t_{1,2}=\frac{a\pm\sqrt{a^2-b^2}}{b}$.

易知,当$x_0>2$时, $x_n>2$;当$0<x_0<2$时, $0<x_n<2$.因此
\[
x_n=\begin{cases}
2\cos \left( \frac{1}{2^n}\arccos \frac{a}{b} \right) ,& 0<a<b;\\
\left( \frac{a+\sqrt{a^2-b^2}}{b} \right) ^{1/2^n}+\left( \frac{a-\sqrt{a^2-b^2}}{b} \right) ^{1/2^n},& 0<b<a.\\
\end{cases}
\]
又因为$\frac{b_{n+1}}{b_n}=\frac{a_{n+1}}{b_{n+1}}=\frac{1}{2}x_{n+1}$,所以,当$0<a<b$时,
\begin{align*}
b_n&=b_0\frac{x_1x_2\cdots x_n}{2^n}=b\cos \frac{\theta}{2}\cos \frac{\theta}{2^2}\cdots \cos \frac{\theta}{2^n}
\\
&=\frac{b\sin \theta}{2^n\sin \frac{\theta}{2^n}},\quad \theta =\arccos \frac{a}{b},n=0,1,2,\ldots
\end{align*}
则
\[a_n=\frac{b\sin \theta}{2^n}\cot \frac{\theta}{2^n}\]
并且\[\lim_{n\rightarrow \infty}a_n=\lim_{n\rightarrow \infty}b_n=\frac{b\sin \theta}{\theta}=\frac{\sqrt{b^2-a^2}}{\arccos \frac{a}{b}}.\]

当$a>b>0$时,
\begin{align*}
x_n&=\left( \frac{a+\sqrt{a^2-b^2}}{b} \right) ^{1/2^n}+\left( \frac{a-\sqrt{a^2-b^2}}{b} \right) ^{1/2^n}
\\
&=e^{\alpha /2^n}+e^{-\alpha /2^n}=2\mathrm{cosh}\left( \frac{\alpha}{2^n} \right),\quad \alpha=\ln\frac{a+\sqrt{a^2-b^2}}{b}
\end{align*}
其中$\mathrm{cosh} x=\frac{1}{2}\left(e^x+e^{-x}\right)$为双曲余弦函数,注意到$\mathrm{sinh}x=2\mathrm{sinh}\frac{x}{2}\mathrm{cosh}\frac{x}{2}$,其中$\mathrm{sinh} x=\frac{1}{2}\left(e^x-e^{-x}\right)$为双曲正弦函数,我们有
\[b_n=b_0\frac{x_1x_2\cdots x_n}{2^n}=b\mathrm{cosh}\frac{\alpha}{2}\mathrm{cosh}\frac{\alpha}{2^2}\cdots \mathrm{cosh}\frac{\alpha}{2^n}=\frac{b\mathrm{sinh}\alpha}{2^n\mathrm{sinh}\frac{\alpha}{2^n}},\]
故
\[a_n=\frac{b\mathrm{sinh}\frac{\alpha}{2}}{2^n}\mathrm{coth} \frac{\alpha}{2^n},\]
并且
\[
\lim_{n\rightarrow \infty}a_n=\lim_{n\rightarrow \infty}b_n=\frac{b\mathrm{sinh} \alpha}{\alpha}=\frac{\sqrt{a^2-b^2}}{\ln \frac{a+\sqrt{a^2-b^2}}{b}}.
\]

(1990年匈牙利奥赛)设$a_0=1,a_n=\frac{\sqrt{1+a_{n-1}^2}-1}{a_{n-1}}$,则$a_n>\frac{\pi}{2^{n+2}}$.

设$a_k\geq 0,a_{n+m}\leq a_n+a_m$,则对任意$n\geq m$,均有
\[a_n\le ma_1+\left( \frac{n}{m}-1 \right) a_m.\]
特别地,若$a_1=1,a_n>1(n\geq 2)$且$a_{n+m}\leq a_n+a_m$,则$a_n<n$.

(1989年30届IMO预选题)定义数列$\{a_n\},\{b_n\}$如下: $a_0=\frac{\sqrt{2}} {2},b_0=1$,
\[ a_{n+1}=a_0\sqrt{1-\sqrt{1-a_n^2}},\quad b_{n+1}=\frac{\sqrt{1+b_n^2}-1}{b_n}(n=0,1,2,\ldots)\]
求证:对每一个$n=0,1,2,\ldots$,有不等式
\[2^{n+2}a_n<\pi<2^{n+2}b_n.\]

提示:令$b_{n}=\tan\theta_n$,可求得$\theta_n=\frac{\pi}{2^{n+2}}$,故$b_n=\tan\frac{\pi} {2^{n+2}}$.再利用不等式$\sin x<x<\tan x(0<x<\pi/2)$即可.

(叶军P 283)已知数列$\{a_n\},\{b_n\}$满足$a_0=a,b_0=b(a>0,b>0)$,且
\[\begin{cases}
a_{n+1}=\frac{1}{2}(a_n+b_n)\\
b_{n+1}=\frac{2a_nb_n}{a_n+b_n}.
\end{cases}(n=0,1,2,\ldots)\]
试求$\{a_n\},\{b_n\}$的通项公式.

易证$a_nb_n=a_0b_0=ab(n\geq 0)$以及
\[a_{n+1}=\frac{1}{2}\left( a_n+\frac{ab}{a_n} \right).\]
因此
\[
\frac{a_{n+1}-\sqrt{ab}}{a_{n+1}+\sqrt{ab}}=\left( \frac{a_n-\sqrt{ab}}{a_n+\sqrt{ab}} \right) ^2.
\]
解得
\begin{align*}
a_n&=\frac{\left( a+\sqrt{ab} \right) ^{2^n}+\left( a+\sqrt{ab} \right) ^{2^n}}{\left( a+\sqrt{ab} \right) ^{2^n}-\left( a-\sqrt{ab} \right) ^{2^n}}\sqrt{ab},
\\
b_n&=\frac{\left( a+\sqrt{ab} \right) ^{2^n}-\left( a+\sqrt{ab} \right) ^{2^n}}{\left( a+\sqrt{ab} \right) ^{2^n}+\left( a-\sqrt{ab} \right) ^{2^n}}\sqrt{ab}.
\end{align*}

 

已知数列$\{a_n\},\{b_n\}$满足$a_0=a,b_0=b(a>0,b>0)$,且
\[\begin{cases}
a_{n+1}=\frac{2a_nb_n}{a_n+b_n}\\
b_{n+1}=\sqrt{a_{n+1}b_n}.
\end{cases}(n=0,1,2,\ldots)\]
试求$\{a_n\},\{b_n\}$的通项公式及极限.

由$b_{n+1}=\sqrt{a_{n+1}b_n}$可知
\[a_{n+1}=\frac{b_{n+1}^{2}}{b_n}.\]
由$a_{n+1}=\frac{2a_nb_n}{a_n+b_n}$可知
\[\frac{b_{n+1}^{2}}{b_n}=\frac{2\frac{b_{n}^{2}}{b_{n-1}}b_n}{\frac{b_{n}^{2}}{b_{n-1}}+b_n}=\frac{2b_{n}^{2}}{b_n+b_{n-1}},\]
故
\[
\left( \frac{b_n}{b_{n+1}} \right) ^2=\frac{1}{2}\left( 1+\frac{b_{n-1}}{b_n} \right).
\]
令$x_n=\frac{b_{n-1}}{b_n}$,则$x_{n+1}^2=\frac{1}{2}\left( 1+x_n \right)$且$x_1=\frac{b_0}{b_1}=\sqrt{\frac{a+b}{2a}}$.

若$0<b\leq a$,则$0<x_1\leq 1$,进一步可知$0<x_n\leq 1$.令$x_n=\cos2\theta_n$,则$x_{n+1}=\sqrt{\frac{1}{2}\left( 1+x_n \right)}=\cos\theta_n$,则$\theta_{n+1}=\frac{1}{2}\theta_n$,则
\[\theta _n=\frac{1}{2^n}\arccos \sqrt{\frac{a+b}{2a}},\quad x_n=\cos \left( \frac{1}{2^{n-1}}\arccos \sqrt{\frac{a+b}{2a}} \right).\]
因此
\begin{align*}
b_n&=b_0\Bigg/\prod_{k=1}^n{\cos \left( \frac{1}{2^{k-1}}\arccos \sqrt{\frac{a+b}{2a}} \right)}=b\Bigg/\frac{\sin \left( \text{2}\arccos \sqrt{\frac{a+b}{2a}} \right)}{2^n\sin \left( \frac{1}{2^{n-1}}\arccos \sqrt{\frac{a+b}{2a}} \right)}
\\
&=\frac{2^n\sin \left( \frac{1}{2^{n-1}}\arccos \sqrt{\frac{a+b}{2a}} \right)}{\sin \left( \text{2}\arccos \sqrt{\frac{a+b}{2a}} \right)}b,
\end{align*}
而
\[
a_n=\frac{b_{n}^{2}}{b_{n-1}}=\frac{2^n\tan \left( \frac{1}{2^{n-1}}\arccos \sqrt{\frac{a+b}{2a}} \right)}{\sin \left( \text{2}\arccos \sqrt{\frac{a+b}{2a}} \right)}b.
\]
特别地,若$a=2\sqrt{3},b=3$,利用$\arccos\frac{\sqrt{6}+\sqrt{2}}{4}=\frac{\pi}{12}$可知
\[
a_n=3\cdot 2^{n+1}\tan \frac{\pi}{3\cdot 2^{n+1}},\qquad b_n=3\cdot 2^{n+1}\sin \frac{\pi}{3\cdot 2^{n+1}}.
\]
此时$\lim_{n\rightarrow \infty}a_n=\lim_{n\rightarrow \infty}b_n=\pi$.

若$b\geq a>0$,则令$x_n=\frac{1}{2}\left( t_n+\frac{1}{t_n} \right)$,则
\[x_{n+1}=\sqrt{\frac{1}{2}\left( 1+x_n \right)}=\frac{1}{2}\left( \sqrt{t_n}+\frac{1}{\sqrt{t_n}} \right),\]
因此$t_{n+1}=\sqrt{t_n}$,由$x_1=\frac{1}{2}\left( t_1+\frac{1}{t_1} \right) =\sqrt{\frac{a+b}{2a}}$可知
\[t_1=\left( \sqrt{\frac{a+b}{2a}}\pm\sqrt{\frac{b-a}{2a}} \right) ^{1/2^{n-1}}.\]
因此
\[
x_n=\frac{1}{2}\left[ \left( \sqrt{\frac{a+b}{2a}}+\sqrt{\frac{b-a}{2a}} \right) ^{\text{1/}2^{n-1}}+\left( \sqrt{\frac{a+b}{2a}}-\sqrt{\frac{b-a}{2a}} \right) ^{\text{1/}2^{n-1}} \right],
\]
则
\begin{align*}
b_n&=\frac{2^{n-1}ab}{\sqrt{b^2-a^2}}\left[ \left( \sqrt{\frac{a+b}{2a}}+\sqrt{\frac{b-a}{2a}} \right) ^{\text{1/}2^{n-1}}-\left( \sqrt{\frac{a+b}{2a}}-\sqrt{\frac{b-a}{2a}} \right) ^{\text{1/}2^{n-1}} \right]
\\
a_n&=\frac{2^nab}{\sqrt{b^2-a^2}}\frac{\left( \sqrt{\frac{a+b}{2a}}+\sqrt{\frac{b-a}{2a}} \right) ^{\text{1/}2^{n-1}}-\left( \sqrt{\frac{a+b}{2a}}-\sqrt{\frac{b-a}{2a}} \right) ^{\text{1/}2^{n-1}}}{\left( \sqrt{\frac{a+b}{2a}}+\sqrt{\frac{b-a}{2a}} \right) ^{\text{1/}2^{n-1}}+\left( \sqrt{\frac{a+b}{2a}}-\sqrt{\frac{b-a}{2a}} \right) ^{\text{1/}2^{n-1}}}.
\end{align*}

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(叶军P 310)设$\{a_n\}$为有下列性质的数列:
\[1=a_0\leq a_1\leq a_2\leq\cdots\leq a_n\leq \cdots\tag{1}\]
又$\{b_n\}$是由下式定义的数列
\[b_n=\sum_{k=1}^n{\left( 1-\frac{a_{k-1}}{a_k} \right) \cdot \frac{1}{\sqrt{a_k}}},\quad n=1,2,\ldots\tag{2}\]
证明: (a)对所有$n=1,2,3,\ldots$,有$0\leq b_n<2$;

(b)对$0\leq c<2$的任一$c$,总存在一个具有性质(1)的数列$\{a_n\}$,使得由(2)导出的数列$\{b_n\}$中有无限多个下标$n$满足$b_n>c$.

因为
\begin{align*}
\left( 1-\frac{a_{k-1}}{a_k} \right) \cdot \frac{1}{\sqrt{a_k}}&=\frac{a_{k-1}}{\sqrt{a_k}}\left( \frac{1}{a_{k-1}}-\frac{1}{a_k} \right)
\\
&=\frac{a_{k-1}}{\sqrt{a_k}}\left( \frac{1}{\sqrt{a_{k-1}}}+\frac{1}{\sqrt{a_k}} \right) \left( \frac{1}{\sqrt{a_{k-1}}}-\frac{1}{\sqrt{a_k}} \right)
\\
&=\left( \sqrt{\frac{a_{k-1}}{a_k}}+\frac{a_{k-1}}{a_k} \right) \left( \frac{1}{\sqrt{a_{k-1}}}-\frac{1}{\sqrt{a_k}} \right) <2\left( \frac{1}{\sqrt{a_{k-1}}}-\frac{1}{\sqrt{a_k}} \right),
\end{align*}
所以
\begin{align*}
0&\le b_n=\sum_{k=1}^n{\left( 1-\frac{a_{k-1}}{a_k} \right) \cdot \frac{1}{\sqrt{a_k}}}
\\
&<\sum_{k=1}^n{2\left( \frac{1}{\sqrt{a_{k-1}}}-\frac{1}{\sqrt{a_k}} \right)}<\frac{2}{\sqrt{a_0}}=2.
\end{align*}

令$\frac{1}{\sqrt{a_k}}=d^k$,因为当$0<d<1$时,条件(1)满足,又和式$b_n$中第$k$项是
\[\left( 1-\frac{d^{-2\left( k-1 \right)}}{d^{-2k}} \right) d^k=\left( 1-d^2 \right) d^k,\]
因此
\begin{align*}
b_n&=\sum_{k=1}^n{\left( 1-d^2 \right) d^k}=\left( 1-d^2 \right) \sum_{k=1}^n{d^k}
\\
&=\left( 1-d^2 \right) \frac{d-d^{n+1}}{1-d}=d\left( 1+d \right) \left( 1-d^n \right).
\end{align*}
现在要求对无穷多个$n$, 均有$d\left( 1+d \right) \left( 1-d^n \right)>c$,即
\[d^n<1-\frac{c}{d(1+d)}.\tag{3}\]
对充分大的$n$, $d^n\to 0$,故只需
\[0<1-\frac{c}{d(1+d)},\quad d(1+d)>c.\]
故只需$d(1+d)>2d^2>c$.因此,只需选择$d$,满足$\sqrt{\frac{c}{2}}<d<1$.这时有$d(1+d)>c$,故(3)式右端为正数,因为$d\in (0,1),d^n\to 0$,所以存在充分大的自然数$M$,使得当$n>M$时(3)式成立,于是(b)得证.